ExamBro
ExamBro
JEE Mains · Physics · STD 12 - 5. Magnetism and matter

A short bar magnet placed with its axis at \(30^{\circ}\) with an external field of 800 Gauss, experiences a torque of \(0.016 N . m\). The work done in moving it from most stable to most unstable position is \(\alpha \times 10^{-3} J\). The value of \(\alpha\) is _________ .

  1. A 32
  2. B 64
  3. C 16
  4. D 128
Verified Solution

Answer & Solution

Correct Answer

(B) 64

Step-by-step Solution

Detailed explanation

\( \tau=\mu B~sin~\theta\Rightarrow0.016=\mu\times B\times\frac{1}{2} \) \( \Rightarrow\mu=\frac{0.032}{B} \) \( W_{ext}=U_{f}-U_{i}=\mu B-(-\mu B)=2\mu B \) \( =2\times\frac{0.032}{B}\times B \) \( =0.064~J \)
Same subject
Explore more questions on app